A 3.0 pF capacitor consists of two parallel plates that have surface charge densities of ±1.0 nC/mm².If the potential between the plates is 23.0 kV,find the surface area of one of the plates.

Question

Quizplus · Accepted Answer

The capacitance C is given by C = ε₀(A/d), where A is the area and d is the separation. The charge Q is given by Q = σA, where σ is the surface charge density. Using Q = CV, we find A = C*V/σ. Substituting the given values, A = (3.0 pF * 23.0 kV) / (1.0 nC/mm²) = 69 mm².

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